Structure B was derived from structure A. What was the main reason for incorporating the two ester groups?
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- Please answer the following questions 1) based on the structures of adenine, cytosine, guanine, and thyme. Why does A prefer to bind to T and C prefer to bind to G 2) were the binding energies the same for AT and CG? Explain the difference 3) which of the four know bases is the unknown synthetic base most like in terms of its binding characteristics? What does this indicate about its structure? 4) if the unknown synthetic base was used to study DNA replication, which natural bases would it interact with? 5) is the picture questionThe following structures were introduced as neuromuscular blocking agents. Structure B is derived from structure A. What was the main reason for incorporating additional ester groups? O To enhance hydrogen bonding interactions with the binding site. O To increase the polarity of the drug. O To decrease the stability of the drug O To alter the conformation of the molecule.Small molecules are used as inhibitors of protein action - as drugs. They most often do this by blocking the active site within the protein. Potential drugs can be screened computationally to determine if they are strongly bound to the protein. Figure 1 shows a possible conformation of a candidate drug molecule, 4-bromo-2- carboxymethylamide-pyrrole (abbreviation: BCMAP) at the active site of a protein (abbreviation: PR). Figure 2 shows the full protein structure whilst figure 3 shows a known inhibitor of the protein at the site, overlayed with another calculated conformer of BCMAP. (c) Outline how you would investigate whether BCMAP would be an effective inhibitor for the protein in competition with the molecules that the protein normally targets, and any solvent. Figure 1. 4-bromo-2-carboxymethylamide-pyrrole (BCMAP) (C, N, O, and Br atoms in yellow, blue, red, and brown, respectively) in the active site of the protein structure.
- Prepare a schematic diagram and present it. The Figure should illustrate the interactions made between the key components of total and non-specific binding reactions in drug bindings. In preparing this figure, it should reflect on the role of each of the components of the reaction mixtures, and why the subtraction of non-specific from total binding allows us to calculate specific binding. The Figure Legend should be brief but informative.agent). The type of binding that is disrupted by mercaptoethanol (a reducing A) Peptide bonds B) Trans double bonds C) Cis double bonds D) Disulfide bondsIn kappa and iota carrageenans, gels are formed through double helical formation of two polysaccharide segments via covalent interactions. True False
- A decapeptide which could resist burns was isolated from an endemic plant from Mt. Mayon. Deduce the amino acid sequence of the decapeptide given the following informations. Mercaptoethanol gave two pentapeptides F1 and F2. Chymotrypsin treatment of F1 releases S and a tetrapeptide (C,K,F,M) while F2 gave a dipeptide (M,S) and a tripeptide (C,F,K) F1 when treated with CNBr released a free homoserine lactone and a tetrapeptide. Trypsin treatment of F1 gave a tripeptide and a dipeptide. F2 treated with carboxypeptidase released free M while with trypsin gave free K and a tetrapeptide.Shortly discuss the above antibodie structure? Please discuss at your own words . Discussion should be to the point (specific 5-6 lines).In a gel electropherosis, Glutamate and valine created a double bond. How is that possible?
- A toxin is glycopeptide, X that is composed of an octapeptide and a saccharide Y. Glycopeptide X is composed of a glycan Y and an octapeptide. The octapeptide was cleaved by trypsin, giving 2 tetrapeptides of exactly the same composition except for its N-terminals. When the octapeptide was cleaved by chymotrypsin, it gave a pentapeptide and a tripeptide. DNFB treatment gave DNP-ser. The composition of the octapeptide is lys, ser, trp.and leu. The leu:ser ratio is 4:1. 1. What is the sequence of this octapeptide? Saccharide Y is a tetrasaccharide. To determine its linkages, a person methylated glycopeptide X effectively methylating saccharide Y. Cleavage gave the peptide and the following products: 2,3,4-tri-O-methyl-a-D-glucopyranoside acid, 2,3-di-O-methyl-ß-D- glucopyranoside, 1,3,4-tri-O- methyl-ß-D-glucuronic acid treatments to the isolated tetrasaccharide gave the following results: and 1,4,6-tri-O-methyl-ß-D-N- acetylgalactosamine. Further 1. A. Treatment of saccharide Y with an…The alkaline hydrolysis of pAUGCAGC oligonucleotide produces: O A. Uridine 2'-monophosphate, uridine 3'-monophosphate, cytosine 2'-monophosphate O B. Adenosine 2'-monophosphate, adenosine 3'-monophosphate, adenosine 21,5'-bisphosphate OC. Guanosine 2'-monophosphate, guanosine 3'-monophosphate, cytosine 3'-monophosphate O D. Cytidine 3'-monophosphate, guanosine 2'-monophosphate, adenine 2'-monophosphate O E. Adenine 3,5'-bisphosphate, guanine 2,5'-bisphosphate, uridine 2'-monophosphate O F. Uridine 2'-monophosphate, uridine 3'-monophosphate, guanine 3'-monophosphateASAP. NOTE: ANSWER ALL THE SUBQUESTIONS TO RECEIVE A LIKE. THANK YOU. CHOOSE THE ANSWER FROM THE FOLLOWING CHOICES: Phosphodiester Bond Glycosidic Bond Ester Bond 1.1. It is the bond present in nucleotides which combines the nitrogenous base and the pentose sugar together. 1.2. It is the bond present in nucleotides which joins the phosphate group and pentose sugar together. 1.3. It is the bond present between nucleotides, combining them together to form DNA and RNA strand.